Prove that $k=0$ is the only $constant$ so $(\sum\limits_{cyc}a^{3}-\sum\limits_{cyc}a^{2}b)-k(a-b)(a-c)(b+c)\geqq0$ .

Question

Given three numbers $a, b, c$ so that $a+ b, b+ c, c+ a\geqq 0$. Prove that $k= 0$ is the only $constant$ so that $$(\sum\limits_{cyc}a^{3}- \sum\limits_{cyc}a^{2}b)- k(a- b)(a- c)(b+ c)\geqq 0$$

This inequality could be generalized from the problem as follow for same conditions with that above $$(\sum\limits_{cyc}a^{3}- \sum\limits_{cyc}a^{2}b)\geqq 0$$ $$\because(\sum\limits_{cyc}a^{3}- \sum\limits_{cyc}a^{2}b)= (a- b)(a- c)(a+ c)+ (b+ c)(b- c)^{2}\geqq 0$$ I've found only $0$ by discriminant, I tried to subs $a, b, c$ as same as an old problem but unsuccessfully

Answer 1

Let $a+b=z$, $a+c=y$ and $b+c=x$.

Thus, $x$, $y$ and $z$ are non-negatives and we obtain: $$\sum_{cyc}(x^2y-xyz)\geq kx(x-y)(x-z),$$ which for $x=1$ and $y=z=0$ gives $k\leq0.$

Now, let $z=0$ and $y>x$.

Thus, $$y\geq k(x-y)$$ or $$k\geq\frac{y}{x-y},$$ which gives $$k\geq\max_{y>x\geq0}\frac{y}{x-y}=-1.$$

Now, show that $k=-1$ is valid and show also that for all $-1\leq k\leq0$ your inequality is true.

Answer 2

We can use substitution here to let $a= y+ z- x, b= z+ x- y, c= x+ y- z$ easily to us. We need to prove for $k=constant= -1$ for $x, y, z\geqq 0$ by using a.m.-g.m.-inequality here, then $$a^{3}- a^{2}b+ b^{3}- b^{2}c+ c^{3}- c^{2}a+ (a- b)(a- c)(b+ c)\geqq 0$$ $$\Rightarrow 4x^{3}+ 4xz^{2}- 8x^{2}z+ 4x^{3}+ 4xz^{2}+ 8y^{2}z- 16xyz\geqq 0$$ For all $k\in [-1, 0]$ with $k= constant$ then combining with $a^{3}- a^{2}b+ b^{3}- b^{2}c+ c^{3}- c^{2}a\geqq 0$ $$\begin{align*} \sum\limits_{cyc}(a^{3}- a^{2}b) &+ k(a- b)(a- c)(b+ c)= \\ &= k\sum\limits_{cyc}(a^{3}- a^{2}b)+ k(a- b)(a- c)(b+ c)+ (1- k)\sum\limits_{cyc}(a^{3}- a^{2}b)\geqq 0 \end{align*}$$ So the inequality holds for the condition $k\in [-1, 0]$ . Similarly, we have the following inequality $$k\left ( \sum\limits_{cyc}a^{2}b(a- b) \right )- b(a+ b- c)(a- c)(c- b)\geqq 0$$ for all $k\in [1, 3], k= constant$ . Op from $\lceil$https://math.stackexchange.com/a/3253850/680032$\rfloor$ .